#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <set>
#include <bitset>
#include <utility>
using namespace std;

#define mm(a, n) memset(a, n, sizeof a)
#define mk(a, b) make_pair(a, b)

const double eps = 1e-6;
const int INF = 0x3f3f3f3f;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<LL, LL> PLL;
typedef pair<int, LL> PIL;

inline void quickread() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

const int N = 1e5 + 10;

struct Node{
    int data, next;
}node[N];

int head, n, k;

int L[N];
int ans[N];

inline void solution() {
    cin >> head >> n >> k;
    for (int i = 0; i < n; i ++ ) {
        int idx;
        cin >> idx;
        cin >> node[idx].data >> node[idx].next;
    }

    int cnt = 0;
    int p = head;
    // 直接将所有链的起始地址保存下来
    while (p != -1) {
        L[cnt ++ ] = p;
        p = node[p].next;
    }

    for (int i = 0; i < cnt; i ++ ) 
        ans[i] = L[i];

    for (int i = 0; i < cnt / k; i ++ ) 
        for (int j = 0; j < k; j ++ )
            ans[i * k + j] = L[(i + 1) * k - 1 - j]; 
    
    for (int i = 0; i < cnt - 1; i ++ ) 
        printf("%05d %d %05d\n", ans[i], node[ans[i]].data, ans[i + 1]);
    printf("%05d %d -1\n", ans[cnt - 1], node[ans[cnt - 1]].data);


}

int main() {
    freopen("input.txt", "r", stdin);
    quickread();
    solution();
    return 0;
}